[proofplan] We prove the Poincare inequality by contradiction and compactness. Assuming the inequality fails, we construct a normalised sequence $(v_k)$ in $W^{1,p}(U)$ with unit $L^p$ norm, zero mean, and gradients converging to zero. The Rellich-Kondrachov compactness theorem extracts a subsequence converging strongly in $L^p(U)$ to a limit $v$ that inherits unit norm, zero mean, and zero weak gradient. Since $U$ is connected and $\nabla v = 0$, the function $v$ is constant a.e., and the zero-mean condition forces $v = 0$, contradicting $\|v\|_{L^p} = 1$. [/proofplan] [step:Assume the inequality fails and construct a normalised sequence] Suppose the conclusion is false. Then for every $k \ge 1$, there exists $u_k \in W^{1,p}(U)$ with: \begin{align*} \|u_k - (u_k)_U\|_{L^p(U)} > k \, \|\nabla u_k\|_{L^p(U)}. \end{align*} Define the normalised sequence: \begin{align*} v_k := \frac{u_k - (u_k)_U}{\|u_k - (u_k)_U\|_{L^p(U)}}. \end{align*} By construction, $(v_k)$ satisfies three properties: - **Unit norm:** $\|v_k\|_{L^p(U)} = 1$. - **Zero mean:** $(v_k)_U = 0$, since subtracting the mean $(u_k)_U$ shifts the average to zero, and dividing by a nonzero constant preserves this. - **Vanishing gradient:** $\nabla v_k = \frac{\nabla u_k}{\|u_k - (u_k)_U\|_{L^p(U)}}$, so $\|\nabla v_k\|_{L^p(U)} < \frac{1}{k} \to 0$. [guided] The proof is by contradiction. If no constant $C$ works, then for each integer $k \ge 1$ there exists $u_k \in W^{1,p}(U)$ such that $\|u_k - (u_k)_U\|_{L^p(U)} > k \|\nabla u_k\|_{L^p(U)}$. Note that $u_k - (u_k)_U \ne 0$ (otherwise the left side is zero and the inequality fails), so the denominator in the normalisation below is nonzero. Define the normalised sequence $v_k := (u_k - (u_k)_U) / \|u_k - (u_k)_U\|_{L^p(U)}$. We verify three properties: **Unit $L^p$ norm.** By construction, $\|v_k\|_{L^p(U)} = \|u_k - (u_k)_U\|_{L^p(U)} / \|u_k - (u_k)_U\|_{L^p(U)} = 1$. **Zero mean.** The mean of $u_k - (u_k)_U$ over $U$ is $(u_k)_U - (u_k)_U = 0$. Dividing by a nonzero scalar preserves this: $(v_k)_U = 0$. **Vanishing gradient.** Since $\nabla v_k = \nabla u_k / \|u_k - (u_k)_U\|_{L^p(U)}$ (the constant $(u_k)_U$ drops out under differentiation), we compute: \begin{align*} \|\nabla v_k\|_{L^p(U)} = \frac{\|\nabla u_k\|_{L^p(U)}}{\|u_k - (u_k)_U\|_{L^p(U)}} < \frac{1}{k} \to 0 \quad \text{as } k \to \infty. \end{align*} The inequality $< 1/k$ follows directly from rearranging the contradiction hypothesis $\|u_k - (u_k)_U\|_{L^p} > k \|\nabla u_k\|_{L^p}$. The vanishing-gradient property is the crucial one: it says the $v_k$ are "almost constant" in the Sobolev sense (their $W^{1,p}$ seminorm tends to zero), while simultaneously maintaining unit $L^p$ norm. This tension -- nearly constant but not collapsing to zero -- is what the compactness argument in the next step will exploit. The sequence $(v_k)$ is also bounded in $W^{1,p}(U)$: we have $\|v_k\|_{W^{1,p}} = \|v_k\|_{L^p} + \|\nabla v_k\|_{L^p} < 1 + 1/k \le 2$. This boundedness is the hypothesis needed to apply the Rellich-Kondrachov compactness theorem. [/guided] [/step] [step:Extract a strongly convergent subsequence in $L^p$ via Rellich-Kondrachov] The sequence $(v_k)$ is bounded in $W^{1,p}(U)$: \begin{align*} \|v_k\|_{W^{1,p}(U)} = \|v_k\|_{L^p(U)} + \|\nabla v_k\|_{L^p(U)} < 1 + \frac{1}{k} \le 2. \end{align*} Since $U$ is bounded with $C^1$ [boundary](/page/Boundary), the Rellich-Kondrachov theorem guarantees that the embedding $W^{1,p}(U) \hookrightarrow L^p(U)$ is compact. Therefore there exists a subsequence $(v_{k_j})$ and a function $v \in L^p(U)$ with: \begin{align*} v_{k_j} \to v \quad \text{strongly in } L^p(U). \end{align*} [guided] We need compactness to extract a convergent subsequence. The sequence $(v_k)$ is bounded in $W^{1,p}(U)$: both the function norm $\|v_k\|_{L^p} = 1$ and the gradient norm $\|\nabla v_k\|_{L^p} < 1/k$ are bounded, so \begin{align*} \|v_k\|_{W^{1,p}(U)} = \|v_k\|_{L^p(U)} + \|\nabla v_k\|_{L^p(U)} < 1 + \frac{1}{k} \le 2. \end{align*} The Rellich-Kondrachov compactness theorem states that for a bounded domain $U \subset \mathbb{R}^n$ with $C^1$ boundary and $1 \le p < \infty$, the inclusion $W^{1,p}(U) \hookrightarrow L^p(U)$ is a compact embedding. This means that every bounded sequence in $W^{1,p}(U)$ has a subsequence that converges strongly in $L^p(U)$. All hypotheses of Rellich-Kondrachov are satisfied: $U$ is bounded, open, with $C^1$ boundary, and $1 \le p < \infty$. Applying the theorem to the bounded sequence $(v_k)$, we extract a subsequence $(v_{k_j})$ and a limit $v \in L^p(U)$ such that \begin{align*} v_{k_j} \to v \quad \text{strongly in } L^p(U). \end{align*} Note that strong $L^p$ convergence is essential here — weak convergence alone would not suffice to pass the $L^p$ norm to the limit in the next step. The compactness of the embedding (as opposed to mere continuity) is what upgrades bounded $W^{1,p}$ sequences to strongly convergent $L^p$ subsequences. [/guided] [/step] [step:Show the limit has unit norm, zero mean, and zero weak gradient] We verify three properties of the limit $v$: **Unit norm.** By continuity of the $L^p$ norm under strong convergence: $\|v\|_{L^p(U)} = \lim_{j \to \infty} \|v_{k_j}\|_{L^p(U)} = 1$. **Zero mean.** Since $v_{k_j} \to v$ in $L^p(U)$, convergence holds in $L^1(U)$ as well (by Holder's inequality, since $\mathcal{L}^n(U) < \infty$). Therefore $(v)_U = \lim_{j \to \infty} (v_{k_j})_U = 0$. **Zero weak gradient.** For any test function $\phi \in C_c^\infty(U)$ and any $i \in \{1, \ldots, n\}$: \begin{align*} \int_U v \, \partial_{x_i} \phi \, d\mathcal{L}^n = \lim_{j \to \infty} \int_U v_{k_j} \, \partial_{x_i} \phi \, d\mathcal{L}^n = -\lim_{j \to \infty} \int_U (\partial_{x_i} v_{k_j}) \, \phi \, d\mathcal{L}^n. \end{align*} The first equality uses $v_{k_j} \to v$ in $L^p(U)$ (so $\int v_{k_j} \psi \, d\mathcal{L}^n \to \int v \psi \, d\mathcal{L}^n$ for any $\psi \in L^{p'}(U)$, where $p' = p/(p-1)$ is the conjugate exponent). The second equality uses integration by parts for the [weak derivative](/page/Weak%20Derivative) of $v_{k_j} \in W^{1,p}(U)$. Applying Holder's inequality with exponents $p$ and $p'$: \begin{align*} \left| \int_U (\partial_{x_i} v_{k_j}) \, \phi \, d\mathcal{L}^n \right| \le \|\nabla v_{k_j}\|_{L^p(U)} \, \|\phi\|_{L^{p'}(U)} < \frac{1}{k_j} \, \|\phi\|_{L^{p'}(U)} \to 0. \end{align*} Therefore $\int_U v \, \partial_{x_i} \phi \, d\mathcal{L}^n = 0$ for all $\phi \in C_c^\infty(U)$ and all $i$, which means $\nabla v = 0$ in the weak sense. [guided] We must show the limit $v$ inherits the three properties from the approximating sequence. Each uses a different mode of convergence. **Unit norm:** The $L^p$ norm is continuous with respect to strong $L^p$ convergence. Since $\|v_{k_j}\|_{L^p} = 1$ for all $j$ and $v_{k_j} \to v$ in $L^p$, we get $\|v\|_{L^p} = 1$. **Zero mean:** We need $\int_U v \, d\mathcal{L}^n = 0$. Since $U$ has finite measure, $L^p(U) \hookrightarrow L^1(U)$ (by Holder: $\|f\|_{L^1} \le \mathcal{L}^n(U)^{1/p'} \|f\|_{L^p}$). So $v_{k_j} \to v$ in $L^1$, and the integral is continuous on $L^1$: $(v)_U = \frac{1}{\mathcal{L}^n(U)} \int_U v \, d\mathcal{L}^n = \lim_{j} \frac{1}{\mathcal{L}^n(U)} \int_U v_{k_j} \, d\mathcal{L}^n = 0$. **Zero weak gradient:** This is the most subtle property. We want to show $\nabla v = 0$ weakly, meaning $\int_U v \, \partial_{x_i} \phi \, d\mathcal{L}^n = 0$ for all $\phi \in C_c^\infty(U)$. By the definition of weak derivative for $v_{k_j}$: \begin{align*} \int_U v_{k_j} \, \partial_{x_i} \phi \, d\mathcal{L}^n = -\int_U (\partial_{x_i} v_{k_j}) \, \phi \, d\mathcal{L}^n. \end{align*} The left side converges to $\int_U v \, \partial_{x_i} \phi \, d\mathcal{L}^n$ (since $v_{k_j} \to v$ in $L^p$ and $\partial_{x_i} \phi \in L^{p'}$). The right side tends to zero because $\|\nabla v_{k_j}\|_{L^p} < 1/k_j \to 0$ and $\phi \in L^{p'}$ is fixed. The Holder inequality quantifies this: \begin{align*} \left|\int_U (\partial_{x_i} v_{k_j}) \, \phi \, d\mathcal{L}^n\right| \le \|\partial_{x_i} v_{k_j}\|_{L^p} \|\phi\|_{L^{p'}} \le \frac{\|\phi\|_{L^{p'}}}{k_j} \to 0. \end{align*} [/guided] [/step] [step:Derive a contradiction from connectedness of $U$] Since $U$ is connected and $\nabla v = 0$ a.e. on $U$, the function $v$ is constant $\mathcal{L}^n$-a.e. on $U$. (A function with zero weak gradient on a connected open set is constant a.e. -- this follows from the characterisation of $W^{1,p}$ functions via absolutely continuous representatives on lines.) Let $v = c$ a.e. for some $c \in \mathbb{R}$. The zero-mean property gives: \begin{align*} 0 = (v)_U = \frac{1}{\mathcal{L}^n(U)} \int_U c \, d\mathcal{L}^n = c. \end{align*} Therefore $v = 0$ a.e. on $U$, which gives $\|v\|_{L^p(U)} = 0$. This contradicts $\|v\|_{L^p(U)} = 1$. The contradiction shows that the Poincare constant $C$ must exist. [guided] The final step converts "zero gradient" into "constant function", which the zero-mean condition then forces to be zero, contradicting unit norm. Why does $\nabla v = 0$ imply $v$ is constant? On a connected open set $U$, a $W^{1,p}$ function with zero weak gradient must be constant a.e. The argument proceeds via the ACL (absolutely continuous on lines) characterisation: for a.e. line segment in $U$ parallel to a coordinate axis, the restriction of $v$ to that segment is absolutely continuous with derivative zero a.e., hence constant on that segment. Since $U$ is connected, any two points can be joined by a path, and the constant value must agree across overlapping line segments, giving $v = c$ a.e. on $U$ for some $c \in \mathbb{R}$. Now apply the zero-mean condition. Since $v = c$ a.e., the mean value is: \begin{align*} (v)_U = \frac{1}{\mathcal{L}^n(U)} \int_U c \, d\mathcal{L}^n = c. \end{align*} We proved $(v)_U = 0$ in the previous step, so $c = 0$ and $v = 0$ a.e. on $U$. But this gives $\|v\|_{L^p(U)} = 0$, which contradicts $\|v\|_{L^p(U)} = 1$ established from strong $L^p$ convergence. The contradiction arose from assuming no constant $C$ satisfies the Poincare inequality. Therefore, there exists $C = C(n, p, U) > 0$ such that $\|u - (u)_U\|_{L^p(U)} \le C \|\nabla u\|_{L^p(U)}$ for all $u \in W^{1,p}(U)$. [/guided] [/step]